| Exam Board | Edexcel |
|---|---|
| Module | CP1 (Core Pure 1) |
| Year | 2022 |
| Session | June |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Volumes of Revolution |
| Type | Volume with trigonometric functions |
| Difficulty | Challenging +1.2 This is a structured multi-part question requiring binomial expansion of cos terms using De Moivre's theorem, then applying the result to integrate cos^6. While it involves several steps and Core Pure content (complex numbers), each part is clearly signposted with the algebraic identity provided. The integration itself becomes routine once part (a) is established. More challenging than a basic C3 volume question but less demanding than questions requiring independent insight or proof construction. |
| Spec | 1.08f Area between two curves: using integration4.02q De Moivre's theorem: multiple angle formulae4.08d Volumes of revolution: about x and y axes |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \(\left(z+\frac{1}{z}\right)^6 = 64\cos^6\theta\) | B1 | |
| \(\left(z+\frac{1}{z}\right)^6 = z^6 + 6z^5\!\left(\frac{1}{z}\right) + 15z^4\!\left(\frac{1}{z^2}\right) + 20z^3\!\left(\frac{1}{z^3}\right) + 15z^2\!\left(\frac{1}{z^4}\right) + 6z\!\left(\frac{1}{z^5}\right) + \left(\frac{1}{z^6}\right)\) | M1 | |
| \(= \left[z^6+\frac{1}{z^6}\right] + 6\left[z^4+\frac{1}{z^4}\right] + 15\left[z^2+\frac{1}{z^2}\right] + 20\) | A1 | |
| Uses \(z^n + \frac{1}{z^n} = 2\cos n\theta\): \(64\cos^6\theta = 2\cos6\theta + 12\cos4\theta + 30\cos2\theta + 20\) | M1 | |
| \(32\cos^6\theta = \cos6\theta + 6\cos4\theta + 15\cos2\theta + 10\) | A1* |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \(H = 2\) | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \(\text{vol} = \left\{\frac{1}{2}\right\}\pi\int\left(2\cos^3\!\left(\frac{x}{4}\right)\right)^2 dx\) | B1ft | |
| \(\text{vol} = \{2\pi\}\int\cos^6\!\left(\frac{x}{4}\right)dx = \{2\pi\}\int\frac{1}{32}\left(\cos\!\frac{6x}{4}+6\cos\!\frac{4x}{4}+15\cos\!\frac{2x}{4}+10\right)dx\) | M1 | |
| \(= \{2\pi\}\left[\frac{1}{32}\left(\frac{2}{3}\sin\!\left(\frac{3x}{2}\right)+6\sin(x)+30\sin\!\left(\frac{x}{2}\right)+10x\right)\right]\) | A1 | |
| Applies limits correctly | dM1 | |
| \(= 24.56\) | A1 |
## Question 8:
### Part (a):
| Working/Answer | Mark | Guidance |
|---|---|---|
| $\left(z+\frac{1}{z}\right)^6 = 64\cos^6\theta$ | B1 | |
| $\left(z+\frac{1}{z}\right)^6 = z^6 + 6z^5\!\left(\frac{1}{z}\right) + 15z^4\!\left(\frac{1}{z^2}\right) + 20z^3\!\left(\frac{1}{z^3}\right) + 15z^2\!\left(\frac{1}{z^4}\right) + 6z\!\left(\frac{1}{z^5}\right) + \left(\frac{1}{z^6}\right)$ | M1 | |
| $= \left[z^6+\frac{1}{z^6}\right] + 6\left[z^4+\frac{1}{z^4}\right] + 15\left[z^2+\frac{1}{z^2}\right] + 20$ | A1 | |
| Uses $z^n + \frac{1}{z^n} = 2\cos n\theta$: $64\cos^6\theta = 2\cos6\theta + 12\cos4\theta + 30\cos2\theta + 20$ | M1 | |
| $32\cos^6\theta = \cos6\theta + 6\cos4\theta + 15\cos2\theta + 10$ | A1* | |
### Part (b):
| Working/Answer | Mark | Guidance |
|---|---|---|
| $H = 2$ | B1 | |
### Part (c):
| Working/Answer | Mark | Guidance |
|---|---|---|
| $\text{vol} = \left\{\frac{1}{2}\right\}\pi\int\left(2\cos^3\!\left(\frac{x}{4}\right)\right)^2 dx$ | B1ft | |
| $\text{vol} = \{2\pi\}\int\cos^6\!\left(\frac{x}{4}\right)dx = \{2\pi\}\int\frac{1}{32}\left(\cos\!\frac{6x}{4}+6\cos\!\frac{4x}{4}+15\cos\!\frac{2x}{4}+10\right)dx$ | M1 | |
| $= \{2\pi\}\left[\frac{1}{32}\left(\frac{2}{3}\sin\!\left(\frac{3x}{2}\right)+6\sin(x)+30\sin\!\left(\frac{x}{2}\right)+10x\right)\right]$ | A1 | |
| Applies limits correctly | dM1 | |
| $= 24.56$ | A1 | |\begin{enumerate}
\item (a) Given
\end{enumerate}
$$z ^ { n } + \frac { 1 } { z ^ { n } } = 2 \cos n \theta \quad n \in \mathbb { N }$$
show that
$$32 \cos ^ { 6 } \theta \equiv \cos 6 \theta + 6 \cos 4 \theta + 15 \cos 2 \theta + 10$$
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{f237de57-ed6d-4bea-8bb0-1b4e5b66d7da-22_218_357_653_331}
\captionsetup{labelformat=empty}
\caption{Figure 1}
\end{center}
\end{figure}
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{f237de57-ed6d-4bea-8bb0-1b4e5b66d7da-22_307_824_621_897}
\captionsetup{labelformat=empty}
\caption{Figure 2}
\end{center}
\end{figure}
Figure 1 shows a solid paperweight with a flat base.\\
Figure 2 shows the curve with equation
$$y = H \cos ^ { 3 } \left( \frac { x } { 4 } \right) \quad - 4 \leqslant x \leqslant 4$$
where $H$ is a positive constant and $x$ is in radians.\\
The region $R$, shown shaded in Figure 2, is bounded by the curve, the line with equation $x = - 4$, the line with equation $x = 4$ and the $x$-axis.
The paperweight is modelled by the solid of revolution formed when $R$ is rotated $\mathbf { 1 8 0 } ^ { \circ }$ about the $x$-axis.
Given that the maximum height of the paperweight is 2 cm ,\\
(b) write down the value of $H$.\\
(c) Using algebraic integration and the result in part (a), determine, in $\mathrm { cm } ^ { 3 }$, the volume of the paperweight, according to the model. Give your answer to 2 decimal places.\\[0pt]
[Solutions based entirely on calculator technology are not acceptable.]\\
(d) State a limitation of the model.
\hfill \mbox{\textit{Edexcel CP1 2022 Q8 [12]}}